The standard solution of FeSCN2+ (prepared by combining 9.00 mL of 0.200 M Fe(NO3)3 w/1.00 mL of 0.0020 M KSCN) has an absorbance of 0.510. If a trial's absorbance is measured to be 0.250 and its initial concentration of SCN– was 0.00050 M, the equilibrium concentration of SCN– will be ?
I'm not sure I understand what you did; however, if I understand correctly, then
(std) = 0.002 M x 1/11 = ?? M
Then A=abc. Let's call b = 1 and evaluate a.
Use A = 0.510; b = 1, c from above, and evaluate the constant a.
Then redo, using the constant of a you just found, A of 0.250, b = 1, and evaluate c, concentration.
Check my thinking.
Sure, let's break down the problem step by step to find the equilibrium concentration of SCN–.
Step 1: Calculate the standard concentration of SCN– (c_standard):
The standard solution is prepared by combining 9.00 mL of 0.200 M Fe(NO3)3 with 1.00 mL of 0.0020 M KSCN. Since the volume of the combined solution is 10.00 mL (9.00 mL + 1.00 mL), we need to calculate the concentration of SCN– in the standard solution.
Concentration of SCN– in the standard solution:
c_standard = (0.0020 M x 1 mL) / 10 mL
c_standard = 0.00020 M
Step 2: Calculate the constant "a" using Beer-Lambert Law (A = abc):
The absorbance (A) of the standard solution is given as 0.510. We need to calculate the value of "a" in A = abc, where b = 1.
0.510 = a x 1 x 0.00020
a = 0.510 / (1 x 0.00020)
a = 2550
So, the constant "a" is 2550.
Step 3: Calculate the equilibrium concentration of SCN– (c_equilibrium):
The absorbance (A) of the trial solution is measured to be 0.250. We need to use the value of "a" calculated in Step 2 to calculate c_equilibrium, where b = 1.
0.250 = 2550 x 1 x c_equilibrium
c_equilibrium = 0.250 / (2550 x 1)
c_equilibrium ≈ 9.80 x 10^(-5) M
Therefore, the equilibrium concentration of SCN– is approximately 9.80 x 10^(-5) M.
Please note that this explanation assumes that the reaction between Fe3+ and SCN– follows a 1:1 stoichiometry.